Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, th... | Correction to your answer: $$2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x^2}}$$
Your approach is going to have trouble because $t$ is a fairly complicated mess.
Note:
$$\frac{x+\sqrt{x^2-1}}{x} = 1+\sqrt{1-\frac{1}{x^2}}\to 2\text{ as } x\to +\infty$$
and
$$\frac{x-\sqrt{x^2-1}}{x}= 1-\sqrt{1-\frac{1}{x^2}}\to 0\text{ as }x\to+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | The final decimal digit of $a^2$ and the final decimal digit of $2b^2$ can't agree unless $a$ and $b$ are both multiples of five, leading to an infinite descent. (Check: the possible last digits of $a^2$ are 0,1,4,5,6,9 and the possible last digits of $2b^2$ are 0,2,8.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 3
} |
Prove that $|U|^2=2[\cosh \sqrt{2 m}- \cos \sqrt{2 m}]$ How can we prove that $|U|^2=2[\cosh \sqrt{2 m}- \cos \sqrt{2 m}]$
When $U=e^{\sqrt{\frac{m}{2}}(1+i)}-e^{-\sqrt{\frac{m}{2}}(1+i)}$
I expand
$e^{\sqrt{\frac{m}{2}}(1+i)}-e^{-\sqrt{\frac{m}{2}}(1+i)}=e^{\sqrt{\frac{m}{2}}}[\cos \sqrt{\frac{m}{2}}+i \sin\sqrt{\... | $|U|^2=U\times U^*$, in which $U^*$ means replacement of $i$ with $-i$.
$|U|^2=U\times U^*=(e^{\sqrt{\frac{m}{2}}(1+i)}-e^{-\sqrt{\frac{m}{2}}(1+i)})\times (e^{\sqrt{\frac{m}{2}}(1-i)}-e^{-\sqrt{\frac{m}{2}}(1-i)})= e^{\sqrt{\frac{m}{2}}(2)}-e^{\sqrt{\frac{m}{2}}(+2i)}-e^{\sqrt{\frac{m}{2}}(-2i)}+e^{\sqrt{\frac{m}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple Absolute Value Inequality
$$|\frac{x-1}{x+3}|\leq 2$$
I solve it as follow:
$|\frac{x-1}{x+3}|\leq 2 \iff -2\leq \frac{x-1}{x+3} \leq 2 \iff -2\leq 1-\frac{4}{x+3} \leq 2 \iff -3\leq \frac{-4}{x+3} \leq 1 \iff \frac{3}{4}\geq \frac{1}{x+3} \geq -\frac{1}{4} \iff \frac{4}{3} \leq x+3 \leq -4 \iff -3+\frac{4}{... | Just break it into sections by sign:
$\left|\frac{x-1}{x+3}\right| =
\left\{
\begin{array}{ll}
\frac{x-1}{x+3} & \mbox{if } x \geq 1 \\
-\frac{x-1}{x+3} & \mbox{if } x \in (-3,1) \\
\frac{x-1}{x+3} & \mbox{if } x \lt -3
\end{array}
\right.$
Then solve the three inequalities based on the sign of $x+3$:
$$x\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Give me a hint to solve this integral $$\int_0^1 \frac{1}{(x^2+x+1)^2} dx$$
I don't have any clue as to where to even start. I am struggling on this one.
Please shed some light by giving me a Hint
| $$\dfrac{d\left[\dfrac{ax^2+bx+c}{x^2+x+1}\right]}{dx}$$
$$=\dfrac{(x^2+x+1)(2ax+b)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
We need $1=(x^2+x+1)(2ax+b)-(ax^2+bx+c)(2x+1)=(a-b)x^2+(a-2c)x+b-c$
$\implies a-b=0,a-2c=0,b-c=1\implies \cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Closed form for the sum $\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$ I want a closed form for the sum $$S=\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$$
where: $a\ne 1<p<n;\quad p\in\mathbb Z$
I know a related identity,
$$\quad\displaystyle \sum_{k=0}^n\left\lfloor\frac{k}{p}\right\rfloor=\left(n+1... | My Solution
Apply the Summation by parts, we have:
\begin{align*} S=\sum_{k=1}^n a^k\left\lfloor\dfrac{k}{p}\right\rfloor&=\dfrac{1}{a-1}\sum_{k=1}^n\left\lfloor\dfrac{k}{p}\right\rfloor\Delta(a^k)
\\ &= \left.\dfrac{1}{a-1}\cdot a^k\left\lfloor\dfrac{k}{p}\right\rfloor\right|_{k=1}^{n+1}-\dfrac{1}{a-1}\sum_{k=1}^n a^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Trigonometry to the 24th power How can I find the value of
$$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$
Specifically, is there some easy method that I am overlooking?
| I would use complex exponentials. We have:
\begin{align}
\sin^{24}\frac{\pi}{24}+\cos^{24}\frac{\pi}{24}&=
\left(\frac{e^{i\frac{\pi}{24}}-e^{-i\frac{\pi}{24}}}{2i}\right)^{24}
+\left(\frac{e^{i\frac{\pi}{24}}+e^{-i\frac{\pi}{24}}}{2}\right)^{24}\\
&=\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} (-1)^k e^{i \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 2
} |
Explanation of Generating Functions Simplification Given the problem:
What is the coefficient of $x^{2005}$ in the generating function $G(x) = \frac{1}{(1-x)^2(1+x)^2}$?
The solution posted starts with:
Let $\frac{1}{(1-x)^2(1+x)^2} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{C}{1+x} + \frac{D}{(1+x)^2}$. Upon simp... | We have
\begin{align}
&\frac A{1-x} + \frac B{(1-x)^2} + \frac C{1+x} + \frac D{(1+x)^2}\\
&=\frac{A(1-x)(1+x)^2 + B(1+x)^2 + C(1-x)^2(1+x) + D(1-x)^2}{(1-x)^2(1+x)^2}\\
&=\frac{A(1+x-x^2-x^3) + B(1+2x+x^2) + C(x^3-x^2-x+1) + D(1-2x+x^2)}{(1-x)^2(1+x)^2}\\
&=\frac{(-A+C)x^3 + (-A+B-C+D)x^2 + (A+2B-C-2D)x + (A+B+C+D)}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse Laplace Transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. I am trying to figure out the inverse laplace transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. Looking at my table of Laplace Transforms in my textbook, it seems that either I must break up this fraction using partial fractions into linear terms or I need to us... | Given:
$$Y(s)=\frac{6s-19}{s^2-6s+13}$$
APPROACH #1:
Completing the square of the denominator gives
$$Y(s)=\frac{6s-19}{(s-3)^2+4}=\frac{6s}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}$$
Algebraic manipulation of each term gives an equivalent equation in a more appropriate form (reasons behind this step will become apparent short... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Combination sum . I want to evaluate the following sum :
$$S(k,k')=\sum_{i} C_{i+k}^k C_{k'-i}^{k}$$ = $$S(k,k')=\sum_{i} \binom{i+k}{k} \binom{k'-i}{k}$$
I tried some steps but couldnt get further than :
$S(k,k')=2*(\sum_{i}^{(k'-k)/2}C_{i+k}^k C_{k'-i}^{k})$
| This yields to snake oil (see e.g. Wilf's "generatingfunctionology").
We want a special case of "Vandermonde on its head":
$$
\sum_k \binom{n - k}{r} \binom{m + k}{s}
$$
We will need the identity:
$$
\sum_k \binom{k}{n} z^k
= \frac{z^n}{(1 - z)^{n + 1}}
$$
and negative binomial coefficients:
$$
\binom{-n}{k}
= (-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant by Gauss Elimination I want to find the determinant by Gauss Elimination Method for the following matrix:
$$\begin{bmatrix}
1 & 2 & 0 & 1\\
0 & -1 &1 & -4\\
1 & 2 & 0 & -2\\
2 & 0& 5 &8
\end{bmatrix}$$
My attempt is
Row3-row1 and Row4-2Row1
$$\begin{bmatrix}
1 & 2 & 0 & 1\\
0 & -1 &1 & -4\\
0 & 0 ... | The operations you performed don't change the determinant; you can also swap two rows, which however multiplies the determinant by $-1$.
Swap rows 3 and 4, you matrix becomes triangular and its determinant is
$$
1\cdot(-1)\cdot 1\cdot(-3)=3
$$
so the given matrix has determinant $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a+b+c$, $a^2+b^2+c^2$, $a^3+b^3+c^3$ are real, then so are a,b,c
Let $a,b,c$ be complex numbers with distinct magnitudes such that $a+b+c$, $a^2+b^2+c^2$, $a^3+b^3+c^3$ are real.
Prove that $a,b,c$ are real numbers as well.
I tried to go for contradiction: WLOG assume $a$ is not real. Since $a+b+c\in \mathbb R$, ... | Hint: As you've shown, $(x-a)(x-b)(x-c)$ has real coefficients. What can be said about the complex roots of a polynomial with real coefficients?
A more geometric proof would be to first show that if any of $a,b,c$ are real, we can replace that value with any real value and it is still true, so you can assume that $|a|<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving recurrence relation without initial condition Any idea on how I can approach this recurrence relation? It is very different to other questions I have encountered where there is only one term of $T(n)$ on the RHS, and the initial condition isn't given as well.
$$T(n)=\frac2n\big(T(0)+T(1)+\ldots+T(n-1)\big)+5n$$... | Use generating functions. Define $t(z) = \sum_{n \ge 0} T(n) z^n$, shift the recurrence and multiply out to get rid of fractions; multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
(n + 1) T(n + 1)
&= \sum_{0 \le k \le n} T(k) + 5 (n + 1)^2 \\
\sum_{n \ge 0} (n + 1) T(n + 1) z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve this indefinite integral with $\arctan$? $$\int \frac{(x^2-1)\;\text{d}x}{(x^4+3x^2+1) \tan^{-1}{\left(\frac{x^2+1}{x}\right)}}$$
We should divide numerator and denominator by $ x^2 $ and put $z=x+\frac{1}{x}$ but I'm still not getting the answer. Please help!
| Consider the integral:
\begin{align}
I(a) = \int \frac{(x^2 - 1) \, dx}{(x^4 + a x^2+ 1) \, \tan^{-1}\left(\frac{x^2+1}{x}\right) }.
\end{align}
The integral can be seen in the form
\begin{align}
I(a) = \int \frac{\left(\frac{x^2-1}{x^2}\right) \, dx}{\left( a -1 + \left( x + \frac{1}{x} \right)^2 \right) \, \tan^{-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Continuously changing Dimensions of a Rectangle The dimension of a rectangle are continousl changing. The width increases at a rate of 3 in/s while the length decreases at the rate of 2 in/s. At one instant the rectangle is a 20-in square. How fast is its area changing 3 seconds later?
The answer is -16.
$$\begin{alig... | "The rectangle is a 20-in square" means that it is a $20\times 20$ square; that is, $L(0) = W(0) = 20$. Three seconds later, $L(3) = 20-3\cdot 2 = 14$ and $W(3) = 20 + 3\cdot 3 = 29$, so that
$$
\frac{dA}{dt}\bigg\lvert_{t=3}
= L(3)\frac{dW}{dt}\bigg\lvert_{t=3} + W(3)\frac{dL}{dt}\bigg\lvert_{t=3}
= 14\cdot 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does one construct the Galois field extension $GF((2^2)^3)$? Looking at past exam question, one asks us to construct a Galois field extension $GF((2^2)^3)$ whenever the primitive irreducible polynomial $p(X) = X^3 + \alpha X^2 + \alpha X + \alpha \in GF(2^2)[X]$ is given. How do you answer this?
| Ok. It is easy to check that $p(X)$ is irreducible. Let $\beta$ be a zero of $p(X)$. Then
$$
\begin{aligned}
\beta^3&=\alpha\beta^2+\alpha\beta+\alpha,\\
\beta^4&=\beta^2+\beta+\alpha^2,\\
\beta^5&=\alpha^2\beta^2+\beta+\alpha,\\
\beta^6&=\alpha^2\beta+1.
\end{aligned}
$$
Now we can observe that $\beta^6+\beta^5+\beta^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Resolve this system: Im tried to resolve this problem:
$$\max\quad f\left( x,y \right) =xy\quad \text{s.a}\quad \begin{cases} x^2 +y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$
Well, i form the lagrangian and the respective gradient, so i had this system to resolve:
$$\begin{cases} y+2\lambda_1 x+\lambda_2=0 \\ x+2\lambda_1 y+... | $z=-(x+y), x^2+y^2+(x+y)^2=1 \implies x^2+y^2+xy=\dfrac{1}{2},x^2+y^2\ge 2xy \implies 3xy \le \dfrac{1}{2} \iff xy\le \dfrac{1}{6}$,
when $x=y \cap x^2=\dfrac{1}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negati... | Since we have $(x-y)^2\ge 0$, we have, for $x\gt 0,y\gt 0$,
$$\begin{align}(x-y)^2\ge 0&\Rightarrow x^2+y^2+2xy\ge 4xy\\&\Rightarrow y(x+y)+x(x+y)\ge 4xy\\&\Rightarrow \frac{1}{x}+\frac 1y\ge\frac{4}{x+y}\end{align}$$
Now set $x=b+c,y=a+d$ and $x=c+d,y=a+b$ to get
$$\frac{1}{b+c}+\frac{1}{a+d}\ge\frac{4}{b+c+a+d}$$and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Limit of a complicated function. Find $$\lim\limits_{x \to 2^{-}} \frac{e^{((x+2)\log 4){\frac{[x+1]}{4}}}-16}{ 4^x -16}$$ where $[x]$ denotes the greatest integer function less than or equal to x.
ATTEMPT:
I tried the following substitution $x=2-h$ and as $x \to 2$, $h \to 0.$ which gave me something like this
$\lim\l... | If we assume that there is a typo in the question (as mentioned in my comment) then (The mentioned typo has been fixed by OP)
We can write the given function as $$f(x) = \dfrac{\exp\left\{\{(x + 2)\log 4\}\cdot\dfrac{[x + 1]}{4}\right\} - 16}{4^{x} - 16}$$ When $x \to 2^{-}$ we can write $[x + 1] = 2$ and then the expr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the Height of the Trapezoid
Problem: The area of a trapezoid is equal to 2 and the sum of his diagonals is equal to 4. Find the trapezoid height.
[QUESTION]: I find a result that implies that the height of the Triangle is not uniquely defined, any help discussing this result or with other solutions is appreciat... | There is a way to find $h$, though quite tedious.
By Heron's formula,
$$ 2=\sqrt{s(s-\hat{A}B)(s-BD)(s-\hat{A}D)},$$
where
$ s=\frac{\hat{A}B+BD+\hat{A}D}{2}=\frac{4+\frac{4}{h}}{2}=2+\frac{2}{h}.$
\begin{eqnarray}
2&=&\sqrt{s(s-\hat{A}B)(s-BD)(s-\hat{A}D)}
\\&=& \sqrt{\left(2+\frac{2}{h}\right)\left(2+\frac{2}{h}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the fourier series of the function Find the fourier series of the function
$g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)}$ for $x \not= k\pi$, and $g(k\pi) = \lim_{x\to k\pi} g(x)$, $(k \in \mathbb{Z})$
| The problem is stated to be
\begin{align}
g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)} \mbox{ for } x \not= k\pi, \mbox{ and } g(k\pi) = \lim_{x\to k\pi} g(x), \hspace{5mm} k \in \mathbb{Z}
\end{align}
Consider the following.
\begin{align}
g(x,t) &= \sum_{n=1}^{\infty} \frac{\sin(nx) \, t^{n}}{\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate an Integral Evaluate:
$$\int_{0}^{\infty}\dfrac{\sin^3(x-\frac{1}{x} )^5}{x^3} dx$$
I've been stumped by this Integral and cannot think of how to evaluate it.
I substituted $\dfrac{1}{x^2}=t \Rightarrow \dfrac{-dt}{2}= \dfrac{dx}{x^3}.$
$$$$However, I can't understand what to do next. $$$$Any help on solvi... | as @achille hint,I have post it
since
$$I=\int_{0}^{+\infty}\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=(\int_{0}^{1}+\int_{1}^{\infty})\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=I_{1}+I_{2}$$
since
$$I_{2}=-\int_{0}^{1}x\sin^3{(\dfrac{1}{x}-x)^5}dx$$
so
$$I=\int_{0}^{1}\left(\dfrac{1}{x^3}-x\right)\sin^3{\left(x-\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how t... | The identity $\binom{x+y}{n} = \sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$ is well-known, it is called the Vandermonde identity. The answer for $\binom{xy}{n}$ can be explained using the notion of a $\lambda$-ring, where here we consider the binomial ring $\mathbb{Z}$ with $\lambda^n(x)=\binom{x}{n}$. The main theorem ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
A tough integral:$\int_0^{+\infty}\left( \frac1{\log(x+1)-\log x}-x-\frac12\right)^2 dx$ I would like to prove the convergence of
$$I=\int_0^{+\infty}\left( \frac1{\log(x+1)-\log x}-x-\frac12\right)^2 dx$$
then obtain a closed form of $I$.
Convergence is ensured by the fact that $x \mapsto f(x)=\left( \frac1{\log(x... | We have:
$$ I = \frac{1}{4}\int_{0}^{+\infty}\left(\frac{1}{2}+\frac{1}{e^t-1}-\frac{1}{t}\right)^2\frac{dt}{\sinh^2(t/2)}$$
where:
$$\frac{1}{2}+\frac{1}{e^t-1}-\frac{1}{t} = \sum_{n\geq 1}\frac{2t}{t^2+4\pi^2 n^2}\tag{1}$$
as well as:
$$ \frac{1}{\sinh^2(t)}=\frac{1}{t^2}+2\sum_{n\geq 1}\frac{(t^2-n^2\pi^2)}{(t^2+n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
system of modular equations. $x\equiv 2\pmod3$
$x\equiv 3\pmod 5$
$x\equiv 7 \pmod{11}$
How can I solve this system for $x$? I've tried all kinds of things using divisibility but no success. Any hints of solutions are greatly appreciated. What value of $x$ satisfies these three equations?
| Here is a hands-on solution:
$x\equiv 2\pmod3$ implies $x=2+3y$.
$x\equiv 3\pmod 5$ then implies $3y\equiv 1 \pmod 5$ and so $y\equiv 2 \pmod 5$ because $2 \cdot 3 \equiv 1 \pmod 5$. This means that $y=2+5z$ and $x=2+6+15z=8+15z$.
$x\equiv 7 \pmod{11}$ then implies $15z \equiv -1 \pmod{11}$ and so $z \equiv -3 \pmod{11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$\lim\limits_{x\rightarrow 0 }\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}$ From the post Evaluating limit (iterated sine function) and some discussions inside, one can collect the
following three limits
\begin{eqnarray*}
\lim_{x\rightarrow 0 }\frac{\sin _{n}x}{x} &=&... | Direct computation
$\def\wi{\subseteq}$
$\sin(x) \in x - \frac{1}{6} x^3 + \frac{1}{120} x^5 - \frac{1}{5040} x^7 + O(x^9)$ as $x \to 0$.
As $x \to 0$ and given function $f$ such that $f(x) \in x + a x^3 + b x^5 + c x^7 + O(x^9)$:
$\sin(f(x))$
$\ \in \left( x + a x^3 + b x^5 + c x^7 + O(x^9) \right)$
$\hphantom{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof the series is finite using following inequality Let
$$a_n=\frac1{\sqrt1}+\frac1{\sqrt 2}+\ldots +\frac1{\sqrt n}-2\sqrt n $$
For the task to prove that $$\tag1-2\le a_n\le -1 $$
I was given the hint
$$\tag2\sqrt{k+1}-\sqrt k<\frac1{2\sqrt k},\qquad \forall k\in\mathbb N^* $$
I managed to prove $(1)$ by other met... | First, let us prove (2), that is
\begin{equation*}
\sqrt{k+1}-\sqrt{k}\leq \frac{1}{2\sqrt{k}},\ for\ k\geq 1.
\end{equation*}
\begin{eqnarray*}
\sqrt{k+1}-\sqrt{k} &=&\frac{\left( \sqrt{k+1}-\sqrt{k}\right) \left( \sqrt{%
k+1}+\sqrt{k}\right) }{\left( \sqrt{k+1}+\sqrt{k}\right) } \\
&=&\frac{1}{\left( \sqrt{k+1}+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the remainder when the sum is divided by $1000$
Find $S \pmod{1000}$ given: $$S = \sum_{n=0}^{2015} n! + n^3 - n^2 + n - 1$$
$$S_0 = 0! + 0 - 0 + 0 -1 = 0$$
$$S_1 = 1! + 1 - 1 + 1 - 1 = 1$$
$$S_2 = 2! + 8 - 4 + 2 - 1 = 7$$
This isn't helping, so:
$n! = n(n-1)(n-2)...(1)$ but that is too complicated.
The $n!$ t... | HINT :
Since $1000=2^3\cdot 5^3$, note that
$$15!\equiv 0\pmod{1000}.$$
Hence, we have
$$\sum_{n=0}^{2015}n!\equiv \sum_{n=0}^{\color{red}{14}}n!\pmod{1000}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
floor function problems There are two parts of my problem.
*
*Given $n$ and $x$, $\lfloor \frac nx \rfloor = q$, what is the maximum
possible value for $x$ such that we obtain the same floor value? i.e.
$\lfloor \frac nx \rfloor = \lfloor \frac{n}{x'}\rfloor = q$, where $x' \geq x $.
I think $x' = \lfloor \frac nq \... | By definition of $\lfloor\cdot\rfloor$, we have
$$\biggl\lfloor \frac{n}{x}\biggr\rfloor = q \iff q \leqslant \frac{n}{x} < q+1$$
for $q\in \mathbb{Z}$. Since we are here dealing with positive integers, the sense of the inequalities is retained when multiplying with $x$ and dividing by $q$ resp. $q+1$, so we have
$$\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Matrix inverses - Why are they derived the way they are? Note that this is not a question of how, but why. I know the mechanics of it, but this is the first thing i've come across that truly seems like magic, rather than a rigorous mathematical process.
There are questions on SE about proofs for the inverse, but niethe... | You can derive it by simultaneous transforming $(A|I)\to(I|A^{-1})$. If $a \ne 0$ and $ad-bc\ne 0$ we get:
$$
\left[
\begin{array}{rr|rr}
a & b & 1 & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & b/a & 1/a & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & b/a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Find the limit of $(2\sin x-\sin 2x)/(x-\sin x)$ as $x\to 0$ without L'Hôpital's rule I wonder how to do this in different way from L'Hôpital's rule:
$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$
Please help me solve this without using L'Hopital's rule.
| Write
\begin{eqnarray*}
\frac{2\sin x-\sin 2x}{x-\sin x} &=&\frac{2\sin x-2\sin x\cos x}{x-\sin x} \\
&=&2\left( \frac{\sin x}{x}\right) \left( \frac{1-\cos x}{x^{2}}\right)
\left( \frac{x^{3}}{x-\sin x}\right)
\end{eqnarray*}
Using standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find the cubic equation of roots $α, β, γ$. Taken from Fitzpatrick $4$ unit course textbook. The question says:
If the cubic equation $\ ax^3+bx^2+cx+d$ has roots $α, β, γ$. Find the cubic equation who's roots are $α^2, β^2, γ^2$
I keep getting a $±$ sign that I can't get rid of. The answer in the back is $x(ax+c)^2=(b... | Hint We can write the given cubic polynomial as
$$ax^3 + bx^2 + cx + d = a(x - \alpha) (x - \beta) (x - \gamma).$$
Expanding the r.h.s. and comparing like terms gives
\begin{align}
b &= -a(\alpha + \beta + \gamma) \\
&\,\,\vdots
\end{align}
(These are essentially Vieta's Formulas.)
Additional hint Similarly, the cub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Straight line is tangent to the curve. The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.
My attempt
Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-... | If line $y=mx+1$ is a tangent to $F(x, y)=0$, where $F(x, y)$ is a polynom of degree $2$, then $F(x, mx+1)=0$ have exactly one solution. Hence, discriminant is zero: $$(6m-2)^2=4\cdot(1+m^2)\cdot5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
a sequence limit with inequality condition Let sequence $\{a_{n}\}$ such
$$\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1}\le\dfrac{\sqrt{n}}{2}\le\sqrt{na_{n}+n}-\sqrt{na_{n}},n\ge 1$$
Find limits
$$\lim_{n\to\infty}n\left(\dfrac{9}{16}-a_{n}\right)$$
I am working on a problem and I am lead to prove the following inequality
$$\sqr... | We have $\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1}\le\dfrac{\sqrt{n}}{2}\le\sqrt{na_{n}+n}-\sqrt{na_{n}}$ so
$(\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1})\dfrac{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}\le\dfrac{\sqrt{n}}{2}\le(\sqrt{na_{n}+n}-\sqrt{na_{n}})\dfrac{\sqrt{na_{n}+n}+\sqrt{na_{n}}}{\sqrt{na_{n}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Uniform Convergence of $\sum_{n=0}^\infty (1-x^2)^2x^n$ on $[0,1]$; subsequent integral Let $a_n = (1-x^2)^2 x^n$. Show that $\sum_{n=0}^\infty a_n$ converges uniformly on $[0,1]$ and deduce that $\int_0^1 \frac{(1-x^2)^2}{1-x} dx = \sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}$.
Attempt: Denoting partial sums by $S_n$, we h... | \begin{eqnarray*}
S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\
&=& (1+x)(1-x^2)(1).
\end{eqnarray*}
the uniform convergence follows from
\begin{eqnarray*}
S_m -S_n&=& \sum_{k=n+1}^m (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x}\xrightarrow[n \to \infty]{} 0 \\
\end{eqnarray*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Verify if 2 functions are inverse to each other According this site, 2 functions $f$ and $g$ are the inverse function of each other, only if both $(f \circ g) (x) = x$ and $(g \circ f) (x) = x$ are true.
Is it really necessary to prove both of them? Can someone please provide an counter example where one is satisfied ... | Try $f(x) = x^2$ and $g(x) = \sqrt{x}$. Then $(f \circ g)(x) = x$, but $(g \circ f)(-1) \neq -1$.
EDIT: If you want $f$ and $g$ to have the same domain, let
$$f(x) = \left\lbrace \begin{array}{rcl}(x + 1)^3 & : & x \le -1 \\ (x - 1)^3 & : & x \ge 1 \\ 0 & : & -1 < x < 1 \end{array} \right.$$
Notice that $f$ definitely ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solve: $-\frac{1}{\sqrt{2}} < \sin \theta + \cos \theta < \frac{1}{\sqrt{2}}$ The question is:
Solve $$-\frac{1}{\sqrt{2}} \lt \sin\theta + \cos\theta < \frac{1}{\sqrt{2}}$$ for values of $\theta$ between $0^\circ$ and $180^\circ$.
I realized that:
$$\begin{align}
-\frac{1}{\sqrt{2}} < \sin\theta + \cosθ &< \frac{1}{... | To solve: $$\sin(2\theta) < -\frac{1}{2}$$
we treat $2\theta$ as a single variable. The domain in your problem is $ 0^\circ < \theta < 180^\circ$. So to get our new domain all we must do is multiply by $2$ to obtain: $0^\circ < 2\theta < 360^\circ$.
Now we must ask ourselves what angles when put through the sine funct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's wh... | Dividing the numerator by $x\sqrt x\sqrt x$ and denominator by $x^2$, we have
$$ \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
= \lim_{x \to \infty} \frac{\sqrt{1+\dfrac1x}\left(\dfrac1{\sqrt x}-\sqrt{2+\dfrac3{\sqrt x}}\right)}{\dfrac7{x^2}-\dfrac6x+4}=\frac{1\cdot(0-\sqrt2)}4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve this exponential equation: $3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$ I tried solving this equation $$3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$$ by taking the log of both sides, but with no results, what do I do? Sorry if this equation is very easy, I couldn't solve it...... | Hint: This reduces to $$3^{2x} + 3\cdot 2^x \cdot 3^x - 4\cdot 2^{2x} = 0$$
If $a=3^x$ and $b=2^x$ then we get $$a^2 + 3ab - 4b^2 = 0.$$
More of a hint: You have $x = \log_3 a = \log_2 b$. So now you have two equations involving $a$ and $b$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof question: Prove that 2^(odd integer) + 5^(odd integer) + 2 is a multiple of 3, and 4^(any integer) + 1 can be expressed as 5n, 5n + 1 or 5n + 2. I had been working on the claim in the above question for sometime now. Statistically speaking, it works. For example:
$$\left(2^1 \right)+\left(5^1\right)+2=9,\\
\left(... | For questions of divisibility, it helps to think in terms of modular arithmetic.
To show something is a multiple of $3$, we can check to see if it is $0 \pmod 3$:
$$2^{2m+1} + 5^{2n+1} + 2 \equiv -1 + 2 + 2 \pmod 3 \equiv 3 \pmod 3 \equiv 0 \pmod 3$$
Note: $5$ raised to any power always ends in $5$, and $5 \equiv 2 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
simplify and evaluate $\frac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$ How do you simplify and evaluate $\dfrac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$?
What is the problem asking?
| Method 1: Notice formula, $$\color{red}{\frac{\tan A-\tan B}{1+\tan A\tan B}=\tan (A-B)}$$ Hence, we have $$\frac{\tan 80^\circ-\tan 20^\circ}{1+\tan 80^\circ\tan 20^\circ}$$$$=\tan (80^\circ-20^\circ)$$ $$ =\tan (60^\circ)$$ $$ =\color{blue}{\sqrt{3}}$$
Method 2: Notice, $$\color{red}{\sin A\cos B-\cos A\sin B=\sin (A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about... | $$z^3+z^*=0.$$
Let $z=x+iy$. Then
$$(x+iy)^3+x-iy=0,$$ so that
$$x+x^3-3xy^2=0,\tag{1}$$ and $$-y+3x^2y-y^3=0.\tag{2}$$
We ignore the trivial solution $x=y=0$ and suppose that $x,y\neq 0$. Then using (1) and (2) we see that $$x^3+x=x\left(1+\frac{y^3+y}{3y}\right).$$ Hence, $$-3xy^2+x\left(1+\frac{y^3+y}{3y}\right)=0.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 6
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Get amount of submatrixes from $a \times b $matrix I was trying to do the following exercise
Given a grid of size $a \times b$, write a formula able t calculate the total number of rectangles contained in this rectangle. All integer sizes and positions are counted.
Examples:
$3 \times 2$ matrix $\Rightarrow 18$
$4 \tim... | Yes, although I find it easier to understand it as $\frac{a(a+1)b(b+1)}{4}$.
Every rectangle can be described by selecting two not necessarily distinct columns (the columns the corners are going to occupy) and two not necessarily distinct rows (the rows the corners are going to occupy).
There are $\binom{a}{2}=\frac{a(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$x^2 + (k-3)x + k = 0$, ranges of k for roots to be of same sign I need some help on the following.
The quadratic that I am dealing with is $x^2 + (k-3)x + k = 0$, and I need to find ranges of values of $k$, for which the roots will have the same sign.
For the roots of $ax^2+bx+c=0$ to have same signs, $a(x^2+ \frac{b}... | We solve the equation: \begin{align} x^2+(k-3)x+k = 0 &\implies x = \frac{3-k \pm \sqrt{k^2-6k+9 - 4\cdot 1\cdot k}}{2} \\ &\implies x = \frac{3-k \pm \sqrt{k^2-10k + 9}}{2}\end{align} For starters, we must have $k^2-10k + 9 = (k-1)(k-9) \geq 0$, which happens if $k \leq 1$ or $k \geq 9$. Recall that two numbers have t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$0<\int_0^\infty\frac{\sin t}{\ln(1+x+t)} dt<\frac{2}{\ln(1+x)}$ This is my first time posting so please excuse me if I don't follow the proper etiquette. This one is a rather hard problem that was assigned to me for my calculus 2 class. Thank you for your help!
For $x > 0$, prove that:
$$0<\int_0^\infty\frac{\sin t}{\... | $$\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t = \int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + t + 2n\pi)} - \frac{1}{\ln(1 + x + t + (2n + 1)\pi)} \right)\right]$$
Note
$$f_n(t) = \frac{1}{\ln(1 + x + 2n\pi + t)} - \frac{1}{\ln(1 + x + (2n + 1)\pi + t)}.$$
The derivative
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to integrate $\int \frac{\arctan x}{x^4} dx$? I have written the integral as $\int x^{-4} \arctan x dx$. Then, by applying by parts, I got $-3\dfrac{\arctan x}{x^3} + 3\int \dfrac{1}{x^3(1 + x^2)} dx$. Now, how can I solve the later integral? Is there any other trick to do this?
| \begin{eqnarray}
\int\frac{\arctan x}{x^4}\,dx&=&-\frac{\arctan x}{3x^3}+\frac13\int\frac{1}{x^3(1+x^2)}\,dx\stackrel{x=1/t}{=}-\frac{\arctan x}{3x^3}-\frac13\int\frac{1}{t^{-3}(1+t^{-2})t^2}\,dt\\[10pt]
&=&-\frac{\arctan x}{3x^3}-\frac13\int\frac{t^3}{1+t^2}\,dt=-\frac{\arctan x}{3x^3}-\frac13\int\frac{t(t^2+1)-t}{1+t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Proving that the sum of the first $2n$ terms of the series $1^2 - 3^2 + 5^2 - \cdots$ is $-8n^2$ by induction
Use mathematical induction to prove the following for the first $2n$ terms of the series $$1^2 - 3^2 + 5^2 - 7^2 + \cdots = -8n^2.$$
As we have odd numbers that are squared we could use $n = 2k-1$.
But the $2... |
As we have odd numbers that are squared we could use n = 2k-1
You are summing the sequence: $\{1^2-3^2, 5^2-7^2, 9^2-11^2, \ldots (4n-3)^2-(4n-1)^2, \ldots\}$
But the 2 sides do not equate for n=1 or 2k-1 (if you set k=1)
See above. The first term is $-8$, and so...
Also need to find the sum to (2n+1) terms.
You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
find the Jordan form and $P$ such that $P^{-1}AP = J$. Consider the matrix $$A = \left(\begin{array}{cccc} -11&0&-9\\32&1&24\\16&0&13 \end{array}\right)$$
I want to find the Jordan form of $A$, with $1$-s at the bottom and the jordan basis, which is $P$ columns such that $P^{-1}AP = J$.
I evaluated the charechteristic ... | Saying that 1 is an eigenvalue for matrix $A$ means that there exist non-zero vectors, $v$, such that $Av= v$ or $(A- I)v= 0$.
Here, that means $$\begin{bmatrix}-12 & 0 & -9 \\ 32 & 0 & 24 \\ 16 & 0 & 12\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}\begin{bmatrix}-12x- 9z \\ 32x+ 24z \\ 16x+ 12z \end{bmatrix}= \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | Not a visual proof, but by the Binomial Theorem, $$(1-x)^{-2}=\sum_0^{\infty}{-2\choose n}(-1)^nx^n$$ Now $${-2\choose n}={-2\cdot-3\cdots(-1-n)\over n!}=(-1)^n(n+1)$$ so $(1-x)^{-2}=\sum_0^{\infty}(n+1)x^n$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 13
} |
Trigonometric equation cos sin and power The problem is $2\cos t - 3\sin^2t +2 = 0$.
I get to $2\cos t -3\sin^2t =-2$
I think that I need to use a trigonometric identity like $\cos(x+y)$ and to divide $2\cos t -3\sin^2t$ with the $\sqrt{2^2+3^2}$
Do you know how to solve this? It should be $\sqrt{2^2 + 3^2}$
| We have, $$2\cos t-3\sin^2 t+2=0$$ $$\implies 2\cos t-3+3\cos^2 t+2=0$$ $$\implies 3\cos^2t+ 2\cos t-1=0$$ Factorizing the expression, we get $$(3\cos t-1)(\cos t+1)=0$$ $$\text{if}\ 3\cos t-1=0 \implies \cos t=\frac{1}{3}\implies \color{blue}{t=2n\pi\pm\cos^{-1}\left(\frac{1}{3}\right)}$$ $$\text{if}\ \cos t+1=0 \impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Checking logarithm inequality.
Which one of the following is true.
$(a.)\ \log_{17} 298=\log_{19} 375 \quad \quad \quad \quad (b.)\ \log_{17} 298<\log_{19} 375\\
(c.)\ \log_{17} 298>\log_{19} 375 \quad \quad \quad \quad (d.)\ \text{cannot be determined} $
$17^{2}=289 $ it has a difference of $9$ and $19^{2}=361... | Let $x=\log_{17}{298}, y=\log_{19}{375}$.
By definition of logarithms,
$17^x = 298$ and $19^y=375$
So
$17^{x-2} = \dfrac{298}{289} = 1 + \dfrac{9}{289} \tag{1}$ and $19^{y-2}=\dfrac{375}{361} = 1 + \dfrac{14}{361} \tag{2}$.
Now take natural logarithms
$(x-2)\ln{17} = \ln(1+\dfrac{9}{289}) \approx \dfrac{9}{289} \tag{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Solve $2\sin^3x + \sin3x +3\sin^2x \cos x + \cos^3x=0$ $2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$
My try:
$$2\sin^3x +3\sin x - 4\sin^3x +\cos x(3\sin^2x+\cos^2x)=0 $$
$$ \cos x(2\sin^2x+1) - 2\sin^3x+3\sin x=0.$$
And then i have no idea.
| Since
$$
\sin 3x=3\cos^2x\sin x-\sin^3x
$$
you actually have
$$
\sin^3x+3\sin^2x\cos x+3\sin x\cos^2x+\cos^3 x=0,
$$
or, using the binomial theorem,
$$
(\sin x+\cos x)^3=0.
$$
I'm sure you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to rewrite $\frac{d}{d(x+c)}$? I would like to know how to rewrite the following equations:
$$
\frac{d (f(x))}{d(x+c)} =0\\
\frac{d^2 (f(x))}{d(x+c)^2} =0\\
$$
Here $x$ is a variable, $c$ is a constant and $f(x)$ is a function of x.
I would also like to know the reasoning behind the answer.
| $$\frac { d }{ dx } \left( f\left( x \right) \left( x+c \right) ^{ -1 } \right) =0\\ \frac { df\left( x \right) }{ dx } \left( x+c \right) ^{ -1 }+f\left( x \right) \frac { d\left( \left( x+c \right) ^{ -1 } \right) }{ dx } =0\\ \frac { df\left( x \right) }{ dx } \frac { 1 }{ \left( x+c \right) } -f\left( x \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$? How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$ ?
Please let me know the fastest method you know of for such type of problems. Is there any way other than finding the left hand ... | Expand $x-2\sqrt{2x-4}$ at $x=4$ to get $\frac{1}{8}(x-4)^2+O((x-4)^3)$. Hence
$$
\sqrt{x-2\sqrt{2x-4}} \approx \sqrt{\frac{1}{8}}|x-4|,
$$
and this should make it clear why $f$ is not differentiable at $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sq... | In general we would guess that $\sqrt a+\sqrt b+\sqrt c$ has eight conjugates, obtainable by toggling signs individually for the surds. However, in this special case we see that $\sqrt a\sqrt b\sqrt c=30$, which cannot change its sign. Hence once we picked the sign of two of the surds, the sign of the third is determin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Differential equation: $Ay'' + By' + Cy = h(x)$ I'm stuck solving the equation $y'' - 3y' + 2y = 2x^3-30$.
The auxiliary equation is $k^2 - 3k + 2 = 0$ where $k_1 = 1, k_2=3$. Thus the general solution is: $$y_g = C_1e^x + C_2e^{3x}$$
Then, I tried to find the particular solution taking into consideration that $h(x) = ... | The equations you should have are: $2A=2 \\ -9A+2B=0 \\ 6A-6B+2C=0 \\ 2B-3C+2D=-30 \\ \text{ Solving this system should give you your solution. }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this recurrence relation: $a_n = 3a_{n-1} + \frac{4^n}{4}$? How can I solve the following recurrence relation?
$$a_n = 3a_{n-1} + \frac{4^n}{4}$$
I know that $a_n^{(h)} = 3a_{n-1}$ and that the characteristic equation is:
$$r-3 = 0$$ and thus:
$$a_n^{(h)} = \alpha_1(3)^n$$
I am struggling with the parti... | Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write your recurrence as:
$$
a_{n + 1} = 3 a_n + 4^n
$$
Multiply by $z^n$, add over $n \ge 0$ to get:
$$
\sum_{n \ge 0} a_{n + 1} z^n
= 3 \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} 4^n z^n
$$
Recognize some sums:
$$
\frac{A(z) - a_0}{z}
= 3 A(z) + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$ and get the same answer?
In the first scenario, the answer is $1/12$
I... | Because $$(A-B)^{-1}\not=A^{-1}-B^{-1}.$$
Note that we have$$(\color{red}{A-B})^{-1}=\frac{1}{\color{red}{A-B}}.$$
For your case, we have
$$\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^{-1}=\dfrac{1}{\dfrac{4}{3}-\dfrac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Help in finding the integral function. Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.
| Remember that $x^3+1 = (x+1)(x^2-x+1)$. That quadratic polynomial cannot be factored using real numbers, since the discriminant ($b^2-4ac\vphantom{\dfrac\int\int}$) is negative. So
$(x^3+1)^3 = (x+1)^3 (x^2-x+1)^3$. That should tell you what the partial fraction decomposition is.
To handle $x^2-x+1$, one (of course)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$ $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$
My try:
$$ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = \sqrt{4}\left(\frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} + \frac{1}{2}\cos{\frac{3x}{2}}\right) = 2\... | We have $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$ $$1+\frac{1}{2}\cos{\frac{3x}{2}} + \frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} = \frac{4}{2}\sin^2{\frac{x}{4}}$$
$$\cos{\frac{3x}{2}}\cos \frac{\pi}{3} +\sin{\frac{3x}{2}}\sin\frac{\pi}{3} =-\left(1-2\sin^2{\frac{x}{4}}\right)$$
$$\cos\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate derivative of integral I tried to calculate the derivative of this integral:
$$\int_{2}^{3+\sqrt{r}} (3 + \sqrt{r}-c) \frac{1}{2}\,{\rm d}c $$
First I took the anti-derivative of the integral:
$$\frac{1}{2}\left(\frac{-c^2}{2}+c\sqrt{r}+3c\right)$$
Then I evaluated the integral:
$$-\frac{1}{2}\left(\frac{3+\... | As it appears the derivative is taken with respect to $r$. The integral in question is given by, and evaluated as, the following:
\begin{align}
I(r) &= \frac{1}{2} \, \int_{2}^{3+\sqrt{r}} (3 + \sqrt{r} - t) \, dt \\
&= \frac{1}{2} \, \left[ (3 + \sqrt{r}) t - \frac{t^{2}}{2} \right]_{2}^{3 + \sqrt{r}} \\
I(r) &= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show divisibility by 7 I was stuck at this question:
Suppose $a^2+b^2=c^2$ for $a,b,c \in \mathbb Z$, and neither $a$ nor $b$ is a multiple of 7. Show that $a^2-b^2$ is a multiple of 7
I tried to write $b^2$ as $c^2-a^2$ then get $a^2-b^2=2a^2-c^2$. But this does not seem to generate the solution.
How to solve proble... | Using Euclid's formula, $a=2mn, b=m^2-n^2$
We have $7\nmid2mn(m^2-n^2)$
Now, $(m^2-n^2)^2-(2mn)^2=m^4+n^4-6m^2n^2\equiv m^4+n^4+m^2n^2\pmod7$
But $(m^2-n^2)(m^4+n^4+m^2n^2)=(m^2)^3-(n^2)^3\equiv1-1\pmod7$ using Fermat's Little Theorem as $(m,7)=(n,7)=1$
$\implies7|(m^4+n^4+m^2n^2)$ as $7\nmid(m^2-n^2)$
Can you take it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry identity $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$ How to prove that $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$?
I tried manupulating the left hand side but ended up in $\frac{\cos x}{\sin^2 x}$.
Can someone show me? Thanks in advance.
| $$\csc(x)\cot(x)=\frac{\cos^3(x)}{\sin^2(x)}+\cos(x)\Longleftrightarrow$$
$$\csc(x)\cot(x)=\frac{\cos^3(x)+\cos(x)\sin^2(x)}{\sin^2(x)}\Longleftrightarrow$$
$$\csc(x)\cot(x)\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\frac{1}{\sin(x)}\cdot\frac{\cos(x)}{\sin(x)}\cdot\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x\rightarrow \infty $ I want to calculate limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x \rightarrow \infty $ without L'Hôpital's rule.
I found this task on the Internet. The answer given by the author is $2$.
$$\lim_{x\rightarrow\infty} x \cdot \sin(\sqrt{x^2+... | It would be absurd not to agree with everything said, but this is how I would approach this task:
$\begin{aligned}\lim\limits_{x\to\infty}\left(x\cdot\sin\left(\sqrt{x^2+3}-\sqrt{x^2+2}\right)\right)&=\lim\limits_{x\to\infty}\frac{\sin\left(x\left(\sqrt{1+\frac3{x^2}}-\sqrt{1+\frac2{x^2}}\right)\right)}{\frac1x}\\&=\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to verify my solution to an separable differential equation? I have this question:
Find the general solution to the separable differential equation
$$
\frac{dy}{dx} = y(1-y).
$$
My attempt is :
$$
\frac{dy}{y(1-y)} = dx
$$
$$
\frac{1}{y(1-y)} = \frac{A}{y}+\frac{B}{(1-y)}
$$
$$
1=A(1-y)+ B y
$$
$$
A=1, B-A=0, so B... | $$
\frac{dy}{y} + \frac{dy}{1-y} = dx\Longrightarrow \frac{dy}{y} - \frac{dy}{y-1} = dx
$$
But now
$$
\ln|y| - \ln|y-1| = x + C\Longrightarrow \ln \left|\frac{y}{y-1}\right| = x + C\Longrightarrow \left|\frac{y}{y-1}\right| = e^{x + C} = C_1e^x
$$
$C_1$ must be positive, but
$$
\frac{y}{y-1} = \pm C_1e^x = De^x,
$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the approximate square root of $(a^4-a^{-4})/(a^2-a^{-2})$ Find the approximate square root of
$$\dfrac{\left( \dfrac{12}{5}\right)^4 - \left( \dfrac{5}{12}\right)^4 }{\left( \dfrac{12}{5}\right)^2 - \left( \dfrac{5}{12}\right)^2}$$
I tried using the formula for $(a^4-b^4)$ and $(a^2-b^2)$. Then cancelled the com... | First, this is straight equal to $\left(\frac{12}{5}\right)^{2} + \left(\frac{5}{12}\right)^{2}$ through a simple cancellation. Now notice that
$$\left(\frac{12}{5}\right)^{2} + \left(\frac{5}{12}\right)^{2} =\left(\frac{12}{5}\right)^{2} + \left(\frac{12}{5}\right)^{-2} = \left(\frac{5}{12}\right)^{2} \left( \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
For what $(a,b) \in R^+$ does $\int^\infty_b (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}})dx$ converge? For what pairs $(a,b) \in R^+$ does this integral converge?
$$
\int\limits^{\infty}_{b} \left (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x... | The integral converges iff $a = b$.
Let's rewrite the expression using big-O notation. So
$$\sqrt{x+a}-\sqrt{x} = x^{1/2}(\sqrt{1+a/x}-1) = x^{1/2}(1+a/2x+O(x^{-2})),$$
Therefore
$$\sqrt{\sqrt{x+a}-\sqrt{x}} = x^{1/4}(a/4x+O(x^{-2}))$$
and similarly
$$\sqrt{\sqrt{x}-\sqrt{x-b}} = x^{1/4}(b/4x+O(x^{-2})).$$
Hence
$$
\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $ (a+b) \cos(x) = a-b$, if $ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$ $ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$ Prove that $ (a+b) \cos(x) = a-b$, if $$ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$$
$$ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$$
I tried solving it with system, ... | As written, the solution to the system of equations is $\sin x = 0$. I think you want
$$\tan x -\sin x={4b\sqrt{ab}\over a^2-b^2}.$$ Then solving the system of equations yields
$$\sin x =\frac12 \left({4a\sqrt{ab}\over a^2-b^2}-{4b\sqrt{ab}\over a^2-b^2}\right)={2\sqrt{ab}\over a+b},$$
which gives $$\cos^2 x = 1 -\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Express $\cosh 2x$ and $\sinh 2x$ in exponential form and hence solve for real values of $x$ the equation:$2 \cosh 2x - \sinh 2x =2$
Express $\cosh 2x$ and $\sinh 2x$ in exponential form and hence solve for real values of $x$ the equation:
$2 \cosh 2x - \sinh 2x =2$
Here is my idea:
$$2 \cosh 2x- \sinh 2x = \frac{... | Hint: i have $$e^{2x}+e^{-2x}-1/2e^{2x}+1/2e^{-2x}=2$$ multiplying by $e^{2x}$ we obtain $$e^{4x}+1-1/2e^{4x}+1/2=2e^{2x}$$ thus we get $$e^{4x}-4e^{2x}+3=0$$
with $$u=e^{2x}$$ we get $$u^2-4u+3=0$$
$u_1=3$ or $u_2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation Algebra: $\sum_{n=2}^\infty \frac{1}{2^{n-1}}$ What rule do we use to get the numerical value of $$\sum_{n=2}^\infty \frac{1}{2^{n-1}}$$
My notes give the answer straight away, without explaining what the steps are.
// Thanks everyone!
| $$\sum_{n=2}^\infty \frac{1}{2^{n-1}}=\frac{1}{2}\sum_{n=2}^\infty \frac{1}{2^{n-2}}=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+...\right)=\frac{1}{2}\frac{1}{1-\frac{1}{2}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the equation of the tangent to $ \ y= \frac{1}{x-4} $ at the point L where $\ x=b \ $ This is what I've done so far:
$$y' \ = \ -\frac{1}{(x-4)^ 2} \ \ \rightarrow \ \
y' (b) \ = \ -\frac{1}{( b-4)^2 } $$
Not sure what to do after this. Am I doing it right?
| Notice, we have $$y=\frac{1}{x-4}$$ substituting $x=b$, we get y-coordinate of the point L as follows $$y=\frac{1}{b-4}$$
The slope of the tangent at general point
$$\frac{dy}{dx}=\frac{-1}{(x-4)^2}$$ Hence, the slope at the point $L\left(b, \frac{1}{b-4} \right)$
$$\frac{dy}{dx}=\frac{-1}{(b-4)^2}$$
The equation of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Estimate for $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ If the triangle ABC has sides $a,b,c$ opposite to the vertices A,B,C respectively and $\Delta$ is the area.The expression $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ is
$(A)\leq16\hspace{1 cm}(B)\geq16\hspace{1 cm}(C)\leq4\hspace{1 cm}(D)\geq4$
Using Herons formula, $\Delta=\sq... | with $a=y+z,b=x+z,c=x+y$ we get
$$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}$$
by AM-GM we have $$(x+y)(x+z)(y+z)\geq 8xyz$$
thus our term above is $$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}\geq \sqrt{8}\sqrt{2}=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad \quad... | Plugging in $a = b = c$, we get
\begin{align*}
3 = \frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}= 3\frac{m - a^2}{2a^2}
&\implies m - a^2 = 2a^2 \\
&\implies m = 3a^2.
\end{align*}
On the other hand, plugging $a = b = c$ into the answers we get
\begin{align*}
&\boldsymbol{(1)} \qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve the trigonometric equation $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$ the question is:
Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.
Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarl... | The equation can be rewritten
$$\sqrt2\sin(x-\frac\pi4)=\sqrt2\sin(2x),$$
hence
$$x-\frac\pi4=2x+2k\pi\text{ or }x-\frac\pi4=\pi-2x+2k\pi,$$
$$x=\frac{24k-3}{12}\pi\text{ or }x=\frac{8k+5}{12}\pi.$$
The second formula covers the first, so that
$$\color{green}{x=\frac{8k+5}{12}\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2 \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$
which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after t... | Starting from what you ought to have:
$(z−(2+\frac{3}{2}i))^2 = -7-i + (2+\frac{3}{2}i)^2$.
we get:
$(z−(2+\frac{3}{2}i))^2 = -7-i + (4+6i-\frac{9}{4}) = \frac{-21+20i}{4} = (\frac{2+5i}{2})^2$.
The last step I obtained by guessing. If you want a systematic way to find the square root of a complex number if it is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How compute $\cos(5\theta)$ and $\sin(5\theta)$? I would like to compute $\cos(5\theta)$ and $\sin(5\theta)$. I can use the formula $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ and $\sin(a+b)=\sin(a)\cos(b)+cos(a)\sin(b)$ but it's a little bit to long. Is there an other way to compute it ?
| \begin{align}
c_2 &=c^2-s^2, \hspace{5mm} s_2=2cs \\
c_4 &= c_2^2-s_2^2 = (c^2-s^2)^2-4c^2s^2 = c^4-6c^2s^2+s^4,\\
s_4 &= 2c_2s_2 = 4cs(c^2-s^2) = 4c^3s-4cs^3. \\
c_5 &= c_4c-s_4s = (c^4-6c^2s^2+s^4)c-(4c^3s-4cs^3)s = c^5-10c^3s^2+5cs^4,\\ s_5 &= c_4s+s_4c = (c^4-6c^2s^2+s^4)s+(4c^3s-4cs^3)c = 5c^4s-10c^2s^3+s^5.
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of Legendre function I'm currently trying to solve the sum
$$
f(x)=\sum\limits_{n=0}^\infty\frac{x^{n+1}}{n+1}P_n(x),
$$
where $P_n(x)$ is the Legendre function of order n.
I also named the sum $f(x)$ since I'm the solution will be a function of $x$. The first thing that can be noticed is that the function $f(x)$ i... | Let:
$$f(x) = \sum _{n=0}^{\infty} \frac{x^{n+1} \, P_{n}(x)}{n+1}$$
As you said:
$$f^{'}(x) = \sum_{n=0}^{\infty} x^{n} \, P_{n}(x) + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \, P_{n}^{'}(x)$$
And using the generating function you showed, we get:
$$f^{'}(x) = \frac{1}{\sqrt{1-x^{2}}} + \sum_{n=0}^{\infty} \frac{x^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
| Try this:
$$3 \cos x + 2\sin x=\sqrt{13}\left(\frac{3}{\sqrt{13}}\cos x+\frac{2}{\sqrt{13}}\sin x\right)\\
=\sqrt{13}\sin(\arcsin\frac{3}{\sqrt{13}}+x)=1$$
You can try to solve it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Simplifying integral:$\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u$ $$I(t)=\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u $$
where $\alpha$ and $t$ are positive constant.
P.S.I would like to edit this problem, because I fo... | Using
$$x^{2} - \frac{a^{2}}{x^{2}} = \left(x - \frac{i a}{x} \right)^{2} + 2 i a $$
then
\begin{align}
I &= \int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du
= e^{-2 i a} \, \int_{0}^{\infty} e^{- \left(u - \frac{i a}{u}\right)^{2}} \, du
\end{align}
where $4 \sqrt{t} \, a = \alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving limit of difference is $0$: $\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$ I am trying to show that
$$\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$$
and I know intuitively this must be so since the "+1" contribution in the first term becomes negligible as $n... | Hint: If we can set $a=(n+1)^{\frac 17}$ and $b=n^{\frac 17}$, then the expression becomes $$\lim_{n \to \infty} [a-b]=0.$$ It suffices to prove that $$a-b \to 0$$ as $n \to \infty$.
The Binomial Theorem at degree $7$ gives us
$$a^7-b^7=(a-b)(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6). \tag{$\star$}$$
We have
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
Solve the system of equations $\begin{cases} xy-2y-3 &=\sqrt{y-x-1}+\sqrt{y-3x+5} \\ (1-y)\sqrt{2x-y}+2(x-1) &=(2x-y-1)\sqrt{y}. \end{cases}$ Solve the following system of equations ($x,y \in \Bbb R$):
$$\begin{cases}
xy-2y-3 &=\sqrt{y-x-1}+\sqrt{y-3x+5} \\
(1-y)\sqrt{2x-y}+2(x-1) &=(2x-y-1)\sqrt{y}.
\end{cases}$$
I ... | Let me try. One has $y \geq 0$ and $y \leq 2x$.
Then, the second equation can be written as follows:
$$(1-y)\sqrt{2x-y} + (2x-1) -1 - (2x-1)\sqrt{y} + y\sqrt{y} = 0$$
$$(1-\sqrt{y})\left( (1+\sqrt{y})\sqrt{2x-y} + (2x-1) - 1 - \sqrt{y} - y\right) = 0$$
$$(1-\sqrt{y})(\sqrt{y}\sqrt{2x-y}-\sqrt{y} + \sqrt{2x-y} - 1 + (2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation: $(9x^2+6x-8)\sqrt{3x+2}+6x+23=27x^2+3\sqrt{10+3x}$ Solve the equation:
$(9x^2+6x-8)\sqrt{3x+2}+6x+23=27x^2+3\sqrt{10+3x}$
I used wolframalpha.com and got only solution $x=-\dfrac{1}{3}$.
And this is my try:
Condition: $x\ge-\dfrac{2}{3}$.
The equation is equivalent to:
$\quad(3x+1)^2\sqrt{3x+2}+9-... | I think the polynomial that you get by multiplying by the three conjugates is
$$59049x^{10}-118098x^9+164025x^8-236196x^7 -441774x^6+133164x^5+373086x^4-390636x^3 -193203x^2+141054x+50641=0$$
This has four real zeros, including $x=-1/3$, but those would include negative signs for the square-roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand ... | Observe $$\left(x-\frac{1}{3}\right)^2=x^2-\frac{2}{3}x+\frac{1}{9}$$$$=\frac{1}{3}\left(3x^2-2x\right)+\frac{1}{9}.$$
This is almost the original expression, we're just missing a $7$. Then,
$$\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}.$$
Now, use the original equality to simplify.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 4
} |
Show that $\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$.
My problem is to show that
$$\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$$
for all $x\in\Bbb R$.
I was thinking of first finding the max and then show that its less than $1/\pi$. But it is hard to find it. I g... | Note that $f$ is odd. Thus, it suffices to look at the interval $[0, \infty)$.
Lemma: For $x$ > $\pi$, we have $$\sin(x) > \frac{(\pi^2-x^2)x}{\pi^2+x^2}.$$
Proof: See
here.
Thus, for $x > \pi$, we have
$$
\frac{1}x - \frac{\sin(x)}{x^2} < \frac{1}x + \frac{\pi^2-x^2}{x(\pi^2+x^2)} := g(x).$$
For $x > \pi$ (or even... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Convergent/divergent series Is the following series divergent/convergent?
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1... | We should be careful here. It's tempting to group the terms into natural subsums and argue the grouped series converges and leave it at that. But the convergence of a grouped series converges does not imply the original series converges in general. The obvious example being $1 + (-1) + 1 + (-1) +\cdots.$ Here's an exa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim\limits_{x\to\infty}\frac{(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)}{\left(1+x^{\frac{n+1}{2}}\right)^n}$ I have written a sequence $$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)$$ as $$\frac{(x^{n+1}-1)(x^n+1)}{x-1}$$
Now, the limit is $$\lim\limits_{x\to\infty}\frac{(x^{n+1}-1)(x^n+1)}{(x-1)\left(1+x^{\frac{n+1}{2}}\right)^n}$$... | Dividing both numerator and denominator by $x\cdot x^2\cdot \ldots\cdot x^n=x^{\frac{n(n+1)}{2}}$ we have
\begin{align*}
\lim_{x\to\infty}\frac{(1+x)(1+x^2)(1+x^3)...(1+x^n)}{\left(1+x^{
\frac{n+1}{2}}\right)^n}&=\lim_{x\to \infty}\frac{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right)\ldots\left(1+\frac{1}{x^n}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\to\infty}\frac{a^n}{n!}$ First I tried to use integration:
$$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$
$$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$
But I could not express it as a riemann integral. Now I am thinking about s... | Let $x_n=\frac{a^n}{n!}$.
$$\left|\frac{x_{n+1}}{x_n}\right|=\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}=\frac{a^{n+1}n!}{a^n (n+1)!}= \frac{a}{n+1}\underset{n\to \infty }{\longrightarrow }0$$
and thus, by $x_n\to 0$ by Ratio test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Prove that the coefficients of a quadratic function with real roots cannot be in geometric progression
Suppose $$ax^2+bx+c$$ is a quadratic polynomial (where $a$, $b$ and $c$ are not equal to zero) that has real roots. Prove that $a$, $b$, and $c$ cannot be consecutive terms in a geometric sequence.
I tried writing t... | Note:
I added a proof that for odd $n$
the only root is $-1$.
Generalizing
Milo Brandt's answer,
which I thought of before I saw his,
this applies to a polynomial
of any even degree.
If the polynomial is of degree $2n$,
using his argument,
we need to find out
how many real roots
$p(x)
=x^{2n}+x^{2n-1}+...+x+1
$
can hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Factorization of a polynomials in complex number. Factorize this expression:
$$a^2+b^2+c^2-ab-bc-ca.$$
The result is
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$
How I can get $\Omega$ here?What's the approach?
| This is equal to:
$$\frac{a^3+b^3+c^3-3abc}{a+b+c}$$
which you can verify by multiplying your polynomial by $a+b+c$. So let's try to factor $a^3+b^3+c^3-3abc$.
So we know that
$$a+b+c\mid a^3+b^3+c^3-3abc$$
(The vertical line means "is a factor of.")
Now, let $\Omega$ be a cube root of unity (usually denoted by $\omega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
$$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m}... | First problem: We have $n+1$ different doughnuts (labelled $1$ to $n+1$) lined up in a row, and want to choose $m+1$ of them for breakfast. This can be done in $\binom{n+1}{m+1}$ ways. Let us count another way.
Maybe the leftmost doughnut chosen is $1$. There are $\binom{n}{m}$ ways to choose the rest.
Maybe the leftm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
If $f’(x) = \sin x + (\sin4x)(\cos x)$, then $f’(2x^2 + \pi/2) $is? If $$f'(x) = \sin x + \sin4x \cdot \cos x,$$ then $$f'(2x^2 + \pi/2)$$ is?
Given answer: $$4x\cos(2x^2) – 4x\sin(8x^2) \sin(2x^2)$$
I tried and I'm getting the answer as $\cos(2x^2) - \sin(8x^2)\sin(2x^2)$
| $$f\'(x) = \sin x + \sin 4x \cos x$$
We need to find $f\'(2x^2+\pi/2)$.
Now
$$f\'(2x^2+π/2) = f\'(2x^2+π/2).Derivative of (2x^2+π/2)$$
$$ = {\sin (2x^2+π/2) + \sin 4(2x^2+π/2).Cos(2x^2+π/2)}.4x$$
$$ = 4x{\cos(2x^2) - \sin(8x^2).\sin(2x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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For a natural number $b$, $N(b)=$ number of natural numbers $ \ a \ $ such that the equation $x^2+ax+b=0$ has integral roots. For a natural number $b$, $N(b)= $ number of natural numbers $a$ such that the equation $x^2+ax+b=0$ has integral roots. What is the lowest possible value of $N(6)$?
| Rote approach. The solutions to $x^2+ax+6=0$ are $x_1=(1/2)(\sqrt{a^2-24}-a), x_2=(1/2)(-\sqrt{a^2-24}-a)$. First, $\sqrt{a^2-24}$ is rational iff it is integral, so we must have $a=\sqrt{c^2+24}$ for some integral ${c}$. Notice that if $c>12$ this expression can't be an integer, as the difference between consecutive s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| $$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ad+bc)(ac+bd)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum lies in the interval $(a_1,a_2)\cup(a_3,a_4)$
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
positive point of maximum lies in the interval
$(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$
... | The zeroes of $f'$ are $-a+7\pm\sqrt{-14a+58}$. Now determine $a$ so that the zeroes are zero, you'll find $a=\pm3$. From here you'll be able to find the desired intervals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4c... | $$2\sin^2x=2\cos^2x-\sqrt3$$
$$\iff2\sin^2x=2(1-\sin^2x)-\sqrt3$$
$$\iff\sin^2x=\dfrac{2-\sqrt3}4=\dfrac{(\sqrt3-1)^2}{8}$$
As $\sin x<0,\sin x=-\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac12\cdot\dfrac1{\sqrt2}-\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}=\sin\left(\dfrac\pi6-\dfrac\pi4\right)$
$x=n\pi+(-1)^n\left(\dfrac\pi6-\dfrac\pi4\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by ind... | Let $a_n = (1 + 1/n)^n.$
We want to show $a_{n+1} - a_{n} \geq \dfrac{1}{n(n+1)}$ for large $n$.
$\dfrac{a_{n+1}}{a_n} = \left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1}.$
The RHS can be expanded as
$\left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1} = \dfrac{n+1}{n} \times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?
| First recall some elementary algebra that if $A\cdot B = 0$, then $A = 0$ or $B = 0$ or both of them $A = 0, B = 0$. So $x(x^2-2) = 0 \to x = 0$ or $x^2-2 = 0$, and $x^2-2 = 0 \to (x-\sqrt{2})(x+\sqrt{2}) = 0 \to x-\sqrt{2} = 0$ or $x +\sqrt{2} = 0 \to x = \sqrt{2}$ or $x = -\sqrt{2}$. Thus there are $3$ solutions: $x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Finding a function satisfying a certain inequality This is a continuation of this post where I tried to find a function $f(n)$ that would satisfy the induction step of an inductive argument and it was shown that such function does not exist.
Trying to fix the problem I've come up with a stronger inductive argument that... | It seems the following.
Provided $n\ge 2$ put $k=l=\left\lfloor\frac{n-2}3\right\rfloor$ and
$p=\left\lfloor\frac{n-3}3\right\rfloor$. Then
$$\left\lfloor\frac{n-3}3\right\rfloor+f\left(n-1,\left\lfloor\frac{n-3}3\right\rfloor\right)\le f\left(n, \left\lfloor\frac{n-2}3\right\rfloor\right),$$
which should give a low... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
optimize distance from point to plane with Lagrange method I am trying to optimize distance from point to plane using Lagrange multiplier.
Usually for such problems you are given specific point like (1,2,3) in 3D, and then an exact plane which is just the subject of Lagrange. But what I have here doesn't specify values... | I think what you are looking for is kind of this
Distance from point to plane
\begin{equation*}
D(\mathbf{x}_{0},\mathbf{P}) = \frac{\vert \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} \vert}
{\Vert \mathbf{w} \Vert_{2}}
\end{equation*}
Let
\begin{equation*}
\mathbf{P} = \lbrace \mathbf{x} \in \mathbb{R}^{n} / \mathbf{w}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a Polynomial in $x-\frac1x$ Given that $x^n - (1/x^n)$ is expressible as a polynomial in $x - (1/x)$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P(x-(1/x)) = x^5 - (1/x)^5.$
To start, I factored, giving:
$P(x-\frac{1}{x}) = (x-\frac{1}{x})(x^4+\frac{1}{x^4}+x^2+\frac{1}{x^2}... | $$\left(x^m-\dfrac1{x^m}\right)\left(x^2+\dfrac1{x^2}\right)=x^{m+2}-\dfrac1{x^{m+2}}+x^{m-2}-\dfrac1{x^{m-2}}$$
$$\implies F_{m+2}=F_m\left(x^2+\dfrac1{x^2}\right)+F_{m-2}$$ where $F_r=x^r-\dfrac1{x^r}$
Now $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=F_1^2+2$
$$F_3=\left(x-\dfrac1x\right)^3+3\left(x-\dfrac1x\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.